在PowerShell中正确初始化JSON的问题

Question

我很难理解JSON在PowerShell中是如何工作的……我想创建和实现一个JSON，但我认为我做错了什么。

此代码不起作用：

$json = @"
{ 
    "Volume" : [
    ]
}
"@ | ConvertFrom-Json | Select-Object -Expand Volume


foreach ($i in 0..2) {
    $blockvalue = @"
        { "UUID" : "uuid-$($i)",
            "Details" : [ { "Name" : "Name-$($i)" }, { "SVM" : "SVM-$($i) " }, { "LastSnapshot" : "snap-$($i)" }, { "Date" : "date-$($i) " } ] }
"@
    $json += (ConvertFrom-Json -InputObject $blockvalue)
}



$json | Select UUID,@{Name="Volume";E={$_.Details | Select -Expand Name}},@{Name="SVM";E={$_.Details | Select -Expand SVM}},@{Name="Date";E={$_.Details | Select -Expand Date}},@{Name="LastSnapshot";E={$_.Details | Select -Expand LastSnapshot}} | FT

输出：

Method invocation failed because [System.Management.Automation.PSObject] does 
not contain a method named 'op_Addition'.
At line:14 char:5
+     $json += (ConvertFrom-Json -InputObject $blockvalue)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : InvalidOperation: (op_Addition:String) [], Runtime 
   Exception
    + FullyQualifiedErrorId : MethodNotFound
 
Method invocation failed because [System.Management.Automation.PSObject] does 
not contain a method named 'op_Addition'.
At line:14 char:5
+     $json += (ConvertFrom-Json -InputObject $blockvalue)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : InvalidOperation: (op_Addition:String) [], Runtime 
   Exception
    + FullyQualifiedErrorId : MethodNotFound
 

UUID             Volume           SVM             Date            LastSnapshot   
----             ------           ---             ----            ------------   
uuid-0           Name-0           SVM-0           date-0          snap-0

所以我找到了一个“解决方案”，我使用下面的代码，它是工作，但它并不干净(最糟糕的是，它不知道为什么它是工作的)：

$json2 = @"
{ 
    "Volume" : [
        {},
        {}
    ]
}
"@ | ConvertFrom-Json | Select-Object -Expand Volume


foreach ($i in 0..2) {
    $blockvalue = @"
        { "UUID" : "uuid-$($i)",
            "Details" : [ { "Name" : "Name-$($i)" }, { "SVM" : "SVM-$($i) " }, { "LastSnapshot" : "snap-$($i)" }, { "Date" : "date-$($i) " } ] }
"@
    $json2 += (ConvertFrom-Json -InputObject $blockvalue)
}

$json2 | Select UUID,@{Name="Volume";E={$_.Details | Select -Expand Name}},@{Name="SVM";E={$_.Details | Select -Expand SVM}},@{Name="Date";E={$_.Details | Select -Expand Date}},@{Name="LastSnapshot";E={$_.Details | Select -Expand LastSnapshot}} | FT

输出：

UUID             Volume           SVM             Date            LastSnapshot   
----             ------           ---             ----            ------------   
                                                                                 
                                                                                 
uuid-0           Name-0           SVM-0           date-0          snap-0         
uuid-1           Name-1           SVM-1           date-1          snap-1         
uuid-2           Name-2           SVM-2           date-2          snap-2

看看我所期待的：

UUID             Volume           SVM             Date            LastSnapshot   
----             ------           ---             ----            ------------   
uuid-0           Name-0           SVM-0           date-0          snap-0         
uuid-1           Name-1           SVM-1           date-1          snap-1         
uuid-2           Name-2           SVM-2           date-2          snap-2

我相信这很容易解决，但我没有PS的天赋…

提前感谢您的帮助:)

Answer 1

除了您正在使用的脚本的复杂性之外，只需声明一个数组，以便可以向其中添加项：

$json2 = @()

foreach ($i in 0..2) {
    $blockvalue = @"
        { "UUID" : "uuid-$($i)",
            "Details" : [ { "Name" : "Name-$($i)" }, { "SVM" : "SVM-$($i) " }, { "LastSnapshot" : "snap-$($i)" }, { "Date" : "date-$($i) " } ] }
"@
    $json2 += (ConvertFrom-Json -InputObject $blockvalue)
}

$json2 | 
Select-Object UUID,
    @{ Name = "Volume";       Expression = { $_.Details | Select-Object -Expand Name         } },
    @{ Name = "SVM";          Expression = { $_.Details | Select-Object -Expand SVM          } },
    @{ Name = "Date";         Expression = { $_.Details | Select-Object -Expand Date         } },
    @{ Name = "LastSnapshot"; Expression = { $_.Details | Select-Object -Expand LastSnapshot } } |
Format-Table

Answer 2

通常，使用+=来追加数组是一种不好的做法。这种方法会创建一个新的数组。原始数组的内容加上右操作数将被复制到新数组中。对于小任务，这没什么大不了的，但对于较大的数据集，它会导致指数级的性能下降。

对于快速POC或控制台类型以外的任何工作，我建议您避免在阵列上使用+=。最常见的方法，通常也是最快的方法是让PowerShell为您累积输出。对Bernard's fine answer的调整：

$json2 = 
foreach ($i in 0..2) {
 @"
{ "UUID" : "uuid-$($i)",
    "Details" : [ { "Name" : "Name-$($i)" }, { "SVM" : "SVM-$($i) " }, { "LastSnapshot" : "snap-$($i)" }, { "Date" : "date-$($i) " } ] }
"@ | ConvertFrom-Json     
}

$json2 | 
Select-Object UUID,
    @{ Name = "Volume";       Expression = { $_.Details | Select-Object -ExpandProperty Name         } },
    @{ Name = "SVM";          Expression = { $_.Details | Select-Object -ExpandProperty SVM          } },
    @{ Name = "Date";         Expression = { $_.Details | Select-Object -ExpandProperty Date         } },
    @{ Name = "LastSnapshot"; Expression = { $_.Details | Select-Object -ExpandProperty LastSnapshot } } | 
Format-Table

我想建议更多的语法糖；通常您不需要在平面字符串上重复调用Select-Object -ExpandProperty ...。这段代码可能看起来更像

$json2 | 
Select-Object UUID,
    @{ Name = "Volume";       Expression = { $_.Details.Name         } },
    @{ Name = "SVM";          Expression = { $_.Details.SVM          } },
    @{ Name = "Date";         Expression = { $_.Details.Date         } },
    @{ Name = "LastSnapshot"; Expression = { $_.Details.LastSnapshot } } | 
Format-Table

然而，这产生了不同寻常的结果。展开上述属性会产生类似于Name-0、Null、Null、Null的数组。我将其追溯到JSON文本/结构。我不确定这是不是您的意图，但生成的对象结构如下：

@{UUID=uuid-0; Details=System.Object[]}
|- UUID = uuid-0
|  |- Length = 6
|- Details
| |- Details[0] = @{Name=Name-0}
| |  |- Name = Name-0
| |- Details[1] = @{SVM=SVM-1}
| |  |- SVM = SVM-1
| |- Details[2] = @{LastSnapShot=LastSnapShot-2}
| |  |- LastSnapShot = LastSnapShot-2
| |- Details[3] = @{Date=Date-3}
|    |- Date = Date-3

Details属性是一个对象数组，包含一系列更多的对象，每个对象都有一个单独的属性，分别是Name、SVM等。我认为这样做的目的可能是让细节成为一个具有名称、支持向量机、LastSnapShot和日期作为属性的对象。如果这是正确的，JSON的创建可能看起来更像下面，其中只有一组外部的花括号。

$json2 = 
ForEach( $i in 0..2 ) {
@"
{ "UUID" : "uuid-$($i)",
    "Details" : [ { "Name" : "Name-$($i)", "SVM" : "SVM-$($i) ", "LastSnapshot" : "snap-$($i)", "Date" : "date-$($i) " } ] }
"@ | ConvertFrom-Json     
}

$json2 | 
Select-Object UUID,
    @{ Name = "Volume";       Expression = { $_.Details.Name         } },
    @{ Name = "SVM";          Expression = { $_.Details.SVM          } },
    @{ Name = "Date";         Expression = { $_.Details.Date         } },
    @{ Name = "LastSnapshot"; Expression = { $_.Details.LastSnapshot } } | 
Format-Table

在这种情况下，可以使用更简洁且可能更快的属性展开语法。